## Nuclei

1 .(a) Two stable isotopes of lithium ^{6}_{3}Li and ^{7}_{3}Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes ^{10}_{5}B and ^{11}_{5}B Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of . ^{10}_{5}B and ^{11}_{5}B.

(a) Mass of ^{6}_{3}Li lithium isotope , m_{1} =6.01512 u

Mass of ^{7}_{3}Li lithium isotope , m = 7.01600 u

Abundance of ^{6}_{3}Li , n_{1} =7.5 %

Abundance of ^{7}_{3}Li , n_{2 } = 92.5%

The atomic mass of lithium atom is given as:

m = m_{1}n_{1} +m_{2}n_{2} / n_{1}+n_{2}

= 6.01512 ×7.5 +7.01600 ×92.5 / 92.5 + 7.5

= 6.940934 u.

(b) Mass of ^{10}_{5}B boron isotope ^{10}_{5}B , m = 10.01294 u

Mass of boron isotope ^{11}_{5}B , m = 11.00931 u

Abundance of ^{10}_{5}B , n_{1 }= x %

Abundance of ^{11}_{5}B , n_{2 }=(100 -x) %

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is given as: m =m_{1}n_{1} +m_{2}n_{2} / n_{1}+n_{2}

10.811 =10.01294 ×x +11.00931 ×( 100 – x) / x +100 -x

1081.11 = 10.01294 x + 1100.931 – 11.00931 x

∴x = 19.821 /0.99637 =19.89 %

And 100 -x = 80.11%

Hence, the abundance of is ^{10}_{5}B 19.89% and that of ^{11}_{5}B is 80.11%

2:The three stable isotopes of neon: ^{20}_{10}Ne, ^{21}_{10}Ne and ,^{22}_{10}Ne,have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Solution 2:

Atomic mass of ^{20}_{10}Ne , m_{1 }=19.99u

Abundance of ^{20}_{10}Ne , η_{1 = 90.51%}

Atomic mass of ^{21}_{10}Ne , m_{2 } =20.99u

Abundance of ^{21}_{10}Ne, η_{2 }=0.27%

Atomic mass of ^{22}_{10}Ne, m_{3}=21.99u

Abundance of ^{22}_{10}Ne,η_{3} =9.22%

The average atomic mass of neon is given as:

m= m_{1}n_{1} +m_{2}n_{2}+m_{3}n_{3} / η_{1 }+ η_{2 }+ η_{3}

=19.99 ×90.51 + 20.99 ×0.27 +21.99 × 9.22 / 90.51+ 0.27 +9.22

= 20.1771 u

3.Obtain the binding energy (in MeV) of a nitrogen nucleus ( ^{14}_{7}N ) , given m( ^{14}_{7}N) =14.00307u

Atomic mass of ( _{7}N^{14} ) nitrogen, m =14.00307u

A nucleus of _{7}N^{14} nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm=7m_{H} +7m_{n} – m

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron,m_{n} = 1.008665 u

Δm = 7 × 1.007825 + 7×1.008665 – 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV /c^{2}

Δm = 0.11236× 931.5 MeV /c^{2}

Hence, the binding energy of the nucleus is given as:

E_{b} = Δmc^{2}

Where, c = Speed of light

E_{b } = 0.11236 × 931.5 ( MeV /c^{2 }) × c^{2}

=104.66334MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

4.Obtain the binding energy of the nuclei ^{56}_{26}Fe and ^{209}_{83}Bi and in units of MeV from the following data:

m(^{56}_{26}Fe ) = 55.934939 u ^{m}( ^{209}_{83}Bi ) =208.980388 u

Atomic mass of (^{56}_{26}Fe ) , m_{1 }=55.934939 u

(^{56}_{26}Fe ) nucleus has 26 protons and ( 56-26) =30 neutrons

Hence, the mass defect of the nucleus, Δm =26 ×m_{H}+ 30 ×m_{n} – m_{1}

Where,

Mass of a proton, m_{H}= 1.007825 u

Mass of a neutron, m_{n }= 1.008665 u

Δm =26 × 1.007825 + 30 × 1.008665 -55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV /c^{2}

Δm = 0.528461 × 931.5 MeV /c^{2}

The binding energy of this nucleus is given as:

E_{b1} =Δmc^{2}

Where, c = Speed of light

∴ E_{b1} = 0.528461 931.5 ( MeV /C^{2}) × C^{2}

=492.26MeV

Average binding energy per nucleon = 492.26/56 =8.79 MeV

Atomic mass of ^{209}_{83}Bi, m_{2 }=208.980388 u

^{209}_{83}Bi nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δm’ = 83 ×m_{H} +126 ×m_{n} -m_{2}

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron, m_{n }= 1.008665 u

Δm’ =83 ×1.007825 + 126 × 1.008665 – 208.980388

=83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV /c^{2}

Δm’ = 1.760877 ×931.5 MeV /c^{2}

Hence, the binding energy of this nucleus is given as:

E_{b2} =Δm’ c’

= 1.760877 × 931.5 ( MeV /c^{2 }) ×c^{2}

=1640.26MeV

Average binding energy per nucleon = 1640.26/209 = 7.848 MeV

5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of _{29}^{63}Cu atoms (of mass 62.92960 u).

Mass of a copper coin, m’ =3g

Atomic mass of _{29} Cu ^{63 } atom, m = 62. 92960 u

The total number of _{29} Cu ^{63 } atoms in the coin, N = N_{A} ×m’/ Mass number

Where,

N_{A} = Avogadro’s number = 6.023 × 10^{23} atoms /g

Mass number = 63 g

∴N=6.023 × 10^{23 } × 3/63 = 2.868 ×10 ^{22 }

_{29} Cu ^{63 } nucleus has 29 protons and (63-29) 34 neutrons

∴ Mass defect of this nucleus, Δm’ = 29 × m_{H} + 34× m_{n} -m

Where,

Mass of a proton,m_{H}= 1.007825 u

Mass of a neutron,m_{n} =1.008665 u

∴ Δm’ = 29 × 1.007825 +34 × 1.008665 – 62.9296

=0.591935 u

Mass defect of all the atoms present in the coin, ∆ m = 0.591935 × 2.868 × 1022

= 1.69766958 ×10 ^{22 }u

But 1 u = 931.5 MeV /C^{2}

∴ Δm = 1.69766958 × 10^{22 } × 931.5 MeV /C^{2}

Hence, the binding energy of the nuclei of the coin is given as:

E_{b} =Δmc^{2}

= 1.69766958 × 10 ^{22 }×931.5 ( MeV /c^{2}) ×c^{2}

= 1.581 ×10^{25} MeV

But 1 MeV = 1.6 × 10 ^{-13}J

E_{b }= 1.581 × 10^{25 } × 1.6 ×10 ^{-13}

= 2.5296 ×10 ^{12}J

This much energy is required to separate all the neutrons and protons from the given coin,

6.Write nuclear reaction equations for

(i) α – decay of 226 /88 Ra

(ii) α – decay of 242/94 Pu

iii)β – decay of 32/15p

(iv) β – decay of 210/83 Bi

Solution 6:

α is a nucleus of helium(_{ 2}He^{ 4} ) and β is an electron (e^{– } for β^{– } and e^{+} for β^{+}) In every α decay, there is a loss of 2 protons and 2 neutrons. In every β^{+} decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β^{–} decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as:

(i) _{88}Ra ^{226 }→ _{86}Rn^{222 }+ _{2}He ^{4}

(ii)94Pu ^{242 }→ _{92}U^{238 }+ _{2}He ^{4}

_

(iii)15P ^{32 }→ _{16}S^{32 }+ e ^{–}+ v

_

(iv)_{83}B ^{210 }→ _{84}PO^{210 }+ e^{+}+ v

(V) _{6}C ^{11}→ _{5B}^{11 }+ e^{+}+ v

(vi) _{43}Tc^{97} → _{42MO}^{97 }+ e^{+}+ v

(vii) _{54}Xe^{120} + e^{+} → _{53}I^{120} + v

7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N_{0}

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N_{0} remains after decay. Hence, we can write:

N/N_{0} = 3.125% = 3.125/100 =1/32

but N/N_{0} = e ^{-λl}

Where , λ = Decay constant t = time

∴ – λt =1/32

– λt = ln l – ln 32

– λt = 0 -3.4657

t= 3.4657/λ

since λ = 0.693/T

∴ t= 3.466/(0.693/T) ≈ 5T years

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N_{0} remains after decay. Hence, we can write:

N/N_{0 }= 1% =1/100

but N/N_{0 }= e ^{-λlt}

∴e ^{-λlt }=1/100

-λ t = ln 1 – ln 100

-λ t =0 -4.6052

t= 4.6052 /λ

Since, λ = 0.639/T

∴t= 4.6052 / ( 0.693/T) = 6.645 T year

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

8.The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive _{6}C ^{14}present with the stable carbon isotope _{6}C^{12} . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of _{6} C^{14 }, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of _{6}C^{14 }dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Decay rate of living carbon-containing matter, R = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of _{6}C^{14} .T_{1/2} =5730 years

The decay rate of the specimen obtained from the Mohenjodaro site:

R’ = 9 decays/min

Let N’ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant,λ and time, t as:

N’/N =R’/R =e^{-λt}

e^{-λt }=9/15 =3/5

-λt = log _{e} 3/5 = -0.5108

∴t = 0.5108/λ

but λ =0.693/T_{1/2 }=0.693 /5730

∴t=0.5108/(0.693 /5730) = 4223.5years

Hence, the approximate age of the Indus-Valley civilization is 4223.5 years

9. Obtain the amount of _{27}Co^{60 }necessary to provide a radioactive source of 8.0 mCi strength. The half-life of _{27}Co^{60 }is 5.3 years.

The strength of the radioactive source is given as:

dN/dt = 8.0 m Ci

=8 × 10 ^{-3 } ×3.7 ×10 ^{10 }

= 29.6 × 10 ^{7 }decay/s

Where,

N = Required number of atoms

Half-life of _{27}Co^{60 }, T _{1/2 } =5.3 years

= 5.3 × 365 ×24 ×60 ×60

=1.67 ×108 s

For decay constant λ , we have the rate of decay as:

dN/dt = λN

Where

λ= 0.693 /T_{1/2 }=0.693 / 1.67 × 10^{8 } s ^{-1}

∴N= ( 1/λ) ( dN/dt)

=29.6 × 10 ^{7 }/ ( 0.693/1.67 × 10^{8})= 7.133 ×10 ^{16 }atoms

for _{27}Co^{60 }

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Mass of 7.133 10 ^{16} atoms = 60 ×7.133 × 10 ^{16 }/ 6.023 ×10 ^{23 } =7.106 × 10 ^{-6 }g

Hence, the amount of _{27}Co^{60} necessary for the purpose is 7.106 × 10 ^{-6 }g

10 .The half-life of _{38} Sr ^{90 } is 28 years. What is the disintegration rate of 15 mg of this isotope?

Half life of _{38} Sr ^{90 }, t _{1/2 } =28years

=28 ×365 ×24 ×60 ×60

=8.83 ×10 ^{8 }s

Mass of the isotope, m = 15 mg

90 g of ^{90} Sr_{38 }atom contains 6.023 × 10^{23} (Avogadro’s number) atoms.

Therefore, 15 mg of ^{90} Sr_{38 contains:}

6.023 × 10 ^{23 }× 15 × 10 ^{-3 }/90 i,e ., 1.0038 ×10 ^{20 }Number of atomms

Rate of disintegration, dN/dt = λN

Where,

λ = decay constant =0.693 / 8.83 ×10 ^{8 } s^{-1}

∴dN/dt = 0.693 ×1.0038 ×10 ^{20 }/8.83 ×10 ^{8 }=7.878 ×10 ^{10 }atoms/s

Hence, the disintegration rate of 15 mg of the given isotope is 7.878 ×10 ^{10 }atoms/s

11.Obtain approximately the ratio of the nuclear radii of the gold isotope _{ 79} Au^{197} and the silver isotope _{47} Ag ^{107}.

Solution 11:

Nuclear radius of the gold isotope _{79}Au^{197 }=R _{Au}

Nuclear radius of the silver isotope _{47}Ag^{107 }= R _{Ag}

Mass number of gold A_{Au }= 197

Mass number of silver,A_{Ag }=107

The ratio of the radii of the two nuclei is related with their mass numbers as:

(R_{Au} / R_{Ag} ) = (A_{Au} / A_{Ag}) 1/3 = 1.2256

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

12. Find the Q-value and the kinetic energy of the emitted α-particle in the α decay of (a) _{88} Ra ^{226 }and (b) _{86 }Rn ^{220}

Given m ( _{88}Ra ^{226} ) = 226.02540u , m (_{222} Rn^{89} )= 222.01750u,

m(_{86}Rn^{220 }) = 220.01137u,m( _{84}po^{216 })= 216.00189u

(a) Alpha particle decay of _{88}Ra^{ 26} emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction

_{88}Ra^{226} → _{86}Ra^{ 222} Ra + _{2}He^{4}

Q-value of

emitted α-particle = (Sum of initial mass − Sum of final mass) c ^{2} Where, c = Speed of

light It is given that:

m ( _{88}Ra^{226} ) =226.02540 u

m(_{86}Rn^{222}) =222.01750 u

m(_{2}He^{4}) =4.002603 u

Q-value =[ 226.02540 – (222.01750 + 4.002603) ] u C^{2}

=0.005297 u C^{2}

But 1 u = 931.5 MeV / C^{2}

Q= 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α – particle = ( Mass number after decay /Mass number before decay) × Q

=( 222/226 ) ×4.94 =4.85 MeV

b)Alpha particle decay of (_{ 86}Rn ^{220 })

_{ 86}Rn ^{220 } →_{ 84} Po ^{216} +_{ 2}He^{4}

It is given that:

Mass ( _{86}Rn ^{220 }) of = 220.01137u

Mass _{(}_{84} Po ^{216 }) of =216.00189u

∴ Q – value = [ 220.01137 – ( 216.00189+ 4.00260) ] ×931.5

≈ 641 MeV

Kinetic energy of the α-particle ( ( 220 – 4 ) / 220 ) × 6.41

=6.29MeV

13.The radionuclide ^{11}C decays according to _{6}C^{11} →_{ 5} B ^{11 } + e ^{+ } + v : T_{1/2 }= 20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values: m ( _{6}C^{11 }) = 11.011434 U and m (_{ 6}B^{ 11 }) = 11.009305 u

Calculate Q and compare it with the maximum energy of the positron emitted

The given nuclear reaction is:

_{6} C ^{11} → _{5}B^{11} + e^{+} + v

Half life of _{6} C ^{11 } nuclei, T_{1/2} =20.3 min

Atomic masses are:

m(_{ 6}C ^{11} ) =11 .011434 u

m( _{6}B^{11} ) = 11.009305u

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q – value (Δ Q ) of the nuclear masses of the _{6}C ^{11}

Δ Q = [ m’ (_{6}C ^{11 }) – [ m’ (_{6}B^{11} ) + m_{e} ] ] c^{2 }…..(1)

Where,

m_{e} = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 m_{e} in the case of ^{11}C and 5 m_{e} in the case of ^{11}B.

Hence, equation (1) reduces to:

Δ Q = [ m (_{6}C ^{11 }) – m(_{6}B^{11} ) – 2m_{e }] c^{2}

Here,m (_{6}C ^{11 }) and m(_{6}B^{11} ) are the atomic masses.

∴ Δ Q = [ 11.011434 – 11.009305 – 2 ×0.000548] c^{2}

=(0.001033 c^{2 }) u

But 1 u = 931.5 MeV / c^{2}

ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron

14.The nucleus _{10 }Ne^{23} decays by β ^{– }emission. Write down the βdecay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m ( _{10 }Ne^{23} ) =22.994466 u

m(_{11 }Na^{23} ) =22.989770 u

In β ^{– }emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.β ^{–} emission of the nucleus

_

_{10} Ne^{ 23} _{10}Ne^{23} → _{11}Na^{23} + e^{–} + v +Q

It is given that:

Atomic mass m (_{10}Ne^{23}) of = 22.994466 u

Atomic mass m (_{11}Na^{23} ) of = 22.989770u

Mass of an electron , m_{e} = 0.000548 u

Q- value of the given reaction is given as:

Q = [ m(_{10}Ne^{23}) – [ m (_{11}Na^{23} ) + m_{e} ] ] c^{2}

There are 10 electrons in _{10}Ne^{23 } and 11 electrons in _{11}Na^{23 } . Hence, the mass of the electron is cancelled in the Q value equation.

∴ Q = [ 22.994466 – 22.9897770 ] c^{2}

= (0.004696c ^{2}) u

But 1 u= 931.5 MeV /c^{2}

∴Q =0.004696 × 931.5 = 4.374 MeV

The daughter nucleus is too heavy as compared to e^{–} and _{v}^{–} . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV

15.The Q value of a nuclear reaction A + b → C + d is defined by Q = [ m_{A}+ m_{b}− m_{C}− m_{d}]c^{2} where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) _{1} H^{1 } + _{1} H^{3} → _{1}H ^{2 } + _{1}H ^{2 }

(ii) _{6}C^{12 } + _{6} C^{12} →_{ 10}Ne ^{20} +_{ 2}He^{4}

Atomic masses are given to be

m( 1H^{2}) = 2.014102 u

m(_{1} H^{3} ) = 3.016049 u

m(_{6} C^{12}) = 12.000000 u

m(_{ 10}Ne ^{20}) = 19.992439 u

(i) The given nuclear reaction is:

_{1}H^{1} +_{ 1}H^{3} → _{1}H^{2} + _{1}H^{2}

Atomic mass m(_{1}H^{1}) =1.0 07825 u

Atomic mass m (_{1}H^{3} ) = 3.0164049 u

Atomic mass m (_{1}H^{2} ) = 2.014102 u

According to the question, the Q-value of the reaction can be written as:

Q = [ m(_{1}H^{1}) + m (_{1}H^{3} ) – 2m (_{1}H^{2} ) ] c^{2}

= [1.007825+3.016049-2× 2.014102] c^{2}

Q= (-0.00433c^{2}) u

But 1 u = 931.5 MeV /c^{2}

∴Q = – 0.00433×931.5 = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii)The given nuclear reaction is : _{6}C^{12 }+_{6}C^{12} →_{10}Ne^{20}+ _{2}He^{4}

It is given that:

Atomic mass of m(_{6}C^{12}) = 12.0 u

Atomic mass of m( _{10}Ne^{12 }) = 19.992439 u

Atomic mass of m(_{2}He^{4}) = 4.002603 u

The Q-Value of this reaction is given as:

Q- [ 2m (_{6}C^{12}) -m(_{10}Ne^{20}) – m(_{2}He^{4})] c^{2}

= [ 2 ×12.0 – 19.992439 -4.002603 ]c^{2}

= (0.004958 c^{2}) u

=0.004958 × 931.5 = 4.618377 MeV

The positive Q-Value of the reaction shows that the reaction is exothermic.

16.Suppose, we think of fission of a _{26}Fe^{56}nucleus into two equal fragments, _{13}Al ^{28}Is the fission energetically possible? Argue by working out Q of the process. Given m(_{26}Fe^{56}) = 55.93494 u and m(_{13}Al ^{28}) =27.9819 u

The fission of _{26}Fe^{56}can be given as:

_{26}Fe^{56 } →_{ 13}Al^{28}

It is given that:

Atomic mass of m ( _{26}Fe^{56}) = 55.93494 u

Atomic mass of m (_{ 13}Al^{28}) = 27.98191 u

The Q-value of this nuclear reaction is given as:

Q= [ m(_{26}Fe^{56}) -2m (_{ 13}Al^{28}) ] c^{2}

= [55.93494 -2 ×27.98191 ] c^{2}

= (-0.02888 c^{2})u

but 1 u = 931.5 MeV /c^{2}

∴Q = – 0.02888 × 931.5 = -26.902 MeV

The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

17.The fission properties of _{94} Pu ^{239 }are very similar to those of _{92} U^{235 .}The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure_{94} Pu ^{239 }undergo fission?

Average energy released per fission of _{94}Pu^{239 } , E_{av =180 MeV}

Amount of pure _{94}Pu^{239 }, m =1 kg = 1000g

N_{A} = Avogadro number = 6.023 × 10^{23}

Mass number of _{94}Pu^{239 } =239 g

1 mole of _{94}Pu^{239 } contains N_{A} atoms.

∴ gm of _{94}Pu^{239 } contains ( ( N_{A }/ Mass number ) × m ) atoms

= (6.023 × 10 ^{23 }/ 239 ) × 1000 = 2.52 × 10 ^{24 } atoms

∴ Total energy released during the fission of 1 kg of _{94}Pu^{239} is calculated as:

E = E_{av }× 2.52 ×10 ^{24}

= 180 × 2.52 × 10 ^{24 } = 4.536 ×10 ^{26} MeV

Hence, 4.536 ×10 ^{26} MeVis released if all the atoms in 1 kg of pure _{94}Pu^{239 }undergo fission

18.A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much _{92} U^{235} did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of _{92} U^{235} and that this nuclide is consumed only by the fission process.

Half life of the fuel of the fission reactor, t _{1/2 }=5years

=5 × 365 × 24×60×60 s

We know that in the fission of 1 g of _{92}U^{235 } nucleus, the energy released is equal to 200MeV. 1 molw,i,e ., 235 g of _{92}U^{235 } contains 6.023 × 10 ^{23}Atoms

∴ 1 g _{92}U^{235 } 6.023 × 10 ^{23 }/ 235 atoms contains×

The total energy generated per gram of _{92}U^{235 } is calculated as:

E=( 6.023 × 10 ^{23 }/ 235 ) × 200MeV /g

=200 ×6.023 × 10 ^{23 }×1.6 × 10 ^{-19} ×10^{6} /235 = 8.20 ×10^{10} J/g

The reactor operates only 80% of the time.

Hence, the amount of _{92}U^{235 }consumed in 5 years by the 1000 MW fission reactor is calculated as:

(5 × 80 × 60×60×365×24×1000×10^{6 }/ 100 ×8.20 × 10^{10 }) g

≈1538 kg

Initial amount of _{92}U^{235 }= 2 × 1538 =3076kg

19.How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

_{1}H^{2} +_{1}H^{2} → _{2}He^{3} ++n+ 3.27 MeV

The given fusion reaction is:

_{1}H^{2} + _{1}H^{2} → _{2}He ^{3} +n+ 3.27 MeV

Amount of deuterium, m = 2kg

1 mole, i.e., 2 g of deuterium contains 6.023 ×10^{23 }atoms

∴2.0 kg of deuterium contains = (6.023 ×10^{23 }/2 ) 2000 = 6.023 ×10^{26}atoms

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴Total energy per nucleus released in the fusion reaction:

E = (3.27 /2 ) 6.023 ×10^{26} MeV

=( 3.27 /2 ) 6.023 ×10^{26 } × 1.6 × 10 ^{-19 } ×10 ^{6}

=1.576 × 10 ^{14 }J

Power of the electric lamp ,p = 100W =100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

=( 1.576 × 10^{ 14}/ 100 )s

1.576 × 10^{ 14 }/ 100 × 60 ×60 × 24 ×365 = 4.9 × 10 ^{4} years

20.Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

When two deuterons collide head-on, the distance between their centres, d is given as

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2fm =2 ×10 ^{-15} m

d =2 × 10 ^{-15} + 2 ×10^{-15} =4 × 10 ^{-15 }m

Charge on a deuteron nucleus = Charge on an electron =e =1.6 ×10 ^{-19} C

Potential energy of the two-deuteron system

V = e ^{2}/4 πε_{0} d

Where

ε_{0 }=permittivity of free space

1/4 πε_{0 }= 9 ×1o ^{9 }Nm^{2}C^{-2}

∴V = 9 ×10 ^{9 } × (1.6 × 10 ^{19})^{2} /4 ×10^{15 }J

= 9 ×10 9 × (1.6 × 10 ^{-19})^{2} /4 × 10 -15 ×(1.6 × 10 ^{-19}) eV

=360 MeV

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

21.From the relation R = R_{0} A^{1/3}, where R_{0} is a constant and A is the mass number of a nucleus,show that the nuclear matter density is nearly constant (i.e. independent of A).

We have the expression for nuclear radius as:

R =R_{0} A^{1/3}

Where,

R_{0} = Constant.

A = Mass number of the nucleus

Nuclear matter density , ρ = Mass of the nucleus /Volumeof the nucleus

Let m be the average mass of the nucleus.

Henc

e, mass of the nucleus = mA

ρ = mA/(4/3 πR^{3}) = 3mA/4π(R_{0} A ^{1/3})^{3} = 3mA / 4 πR_{0}^{3} A = 3m / 4 πR_{0}^{3}

Hence, the nuclear matter density is independent of A. It is nearly constant

22.For the β^{+}(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K− shell, is captured by the nucleus and a neutrino is emitted)

e ^{+}+ ^{A}_{Z} X → ^{A}_{Z-1} Y +v

Show that if β^{+} emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

Let the amount of energy released during the electron capture process be Q_{1}. The nuclear reaction can be written as:

e ^{+}+ ^{A}_{Z} X → ^{A}_{Z-1} Y +v +Q_{1 …….(i)}

Let the amount of energy released during the positron capture process be Q_{2}. The nuclear

reaction can be written as:

e ^{+}+ ^{A}_{Z} X → ^{A}_{Z-1} Y +v +Q_{2 ……..(ii)}

m_{N} (_{z} X^{A}) = Nuclear mass of _{z} X^{A}

m_{N ( z-1 Y A ) = Nuclear mass of z-1 Y A}

m(_{z} X^{A}) = Atomic mass of _{z} X^{A}

m(_{z-1 Y A}) Atomic mass of _{z-1 Y A }

m_{e} = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q1 = [ m_{N} (_{z} X^{A}) +m_{e} – m_{N} (_{z-1 Y A }) ] c^{2}

= [m(_{z} X^{A}) – Zm_{e} + m_{e} – m (_{z-1 Y A}) + (z-1)m_{e} ] c^{2}

= [ m(_{z} X^{A}) -m(_{z-1 Y A})] c^{2 }……….(iii)

Q-value of the positron capture reaction is given as:

Q_{2 }=[m_{N}(_{z} X^{A}) – m_{N (z-1 Y A) – me ] c2}

= [m(_{z} X^{A}) – Zm_{e } -m (_{z-1 Y A}) + (z-1) m_{e }-m_{e }] _{ c2}

= [m(_{z} X^{A}) – m(_{z-1 Y A}) – 2m_{e }] c^{2 }……(iv)

It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1 > 0, it does not necessarily mean that Q2 > 0.

In other words, this means that if β^{+} emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

23.In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are _{12} Mg ^{24 }(23.98504u), _{12} Mg ^{25 }(24.98584u) and _{12} Mg ^{26 } (25.98259u)The natural abundance of _{12} Mg ^{24 }is 78.99% by mass. Calculate the abundances of other two isotopes.

Average atomic mass of magnesium, m =24.312 u

Mass of magnesium _{12} Mg^{ 24} isotope, m_{1 }= 23.98504 u

Mass of magnesium _{12} Mg ^{25} isotope, m_{2 }= 24.98584 u

Mass of magnesium _{12} Mg ^{26} isotope, m_{3 }= 25.98259 u

Abundance of , _{12} Mg^{ 24} ,n_{1 } η =78.99%

Abundance of , _{12} Mg^{ 25} ,n_{2 } η_{2 }=x%

Hence, abundance of _{12} Mg^{ 26 } , η_{3}= 100 -x – 78.99% = (21.01 – x)%

We have the relation for the average atomic mass as:

m = m_{1}η_{1} + m_{2}η_{2} + m_{3}η_{3 / } η_{1} +η_{2}+η_{3 }

24.312 = 23.98504 × 78.99 + 24.98584 × x + 25.98259 × (21.01 -x) /100

2431.2 = 1894.5783096 + 24.98584x + 545.8942159 – 25.98259 x

0.99675 x = 9.2725255

∴ x ≈ 9.3 %

and 21.01 -x = 11.71%

Hence, the abundance of _{12} Mg ^{25 } is 9.3% and that of_{12} Mg^{ 26} is 11.71%

24.The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei _{20}Ca^{41} and_{13} AI^{27}from the following data:

m(_{20} Ca ^{40}) = 39 .962591 u

m(_{20} Ca ^{41}) = 40.962278 u

m(_{13} AI^{26}) = 25.96895u

For a neutron removal from _{20} Ca ^{41 } nucleus, the corresponding nuclear reaction can be written as:

_{20} Ca ^{41 } → _{20} Ca ^{40 } + _{0}n^{1 }

It is given that:

m(_{20} Ca ^{40 } ) mass = 39.962591 u

m(_{20} Ca ^{41} ) mass =40.962278 u

m( _{0}n^{1 }) mass =1.00866 u

The mass defect of this reaction is given as:

Δ m =m (_{20} Ca ^{40}) + (_{0}n^{1 }) – m (_{20} Ca ^{41})

=39.962591 +1.008665 – 40.962278= 0.008978 u

but 1 u = 931.5 MeV /c^{2}

∴Δm =0.008978 × 931.5 MeV /c^{2}

Hence, the energy required for neutron removal is calculated as:

E =Δmc^{2}

= 0.008978 × 931.5 = 8.363007 MeV

For _{13}Al^{ 27 }, the neutron removal reaction can be written as:

_{13}Al^{ 2}^{7 } → _{13}Al^{ 26 }+ _{0}n^{1 }

It is given that:

m(_{13}Al^{ 2}^{7}) mass= 26.981541 u

m(_{13}Al^{ 26 })mass= 25.986895 u

The mass defect of this reaction is given as:

Δm =m (_{13}Al^{ 26}) + m(_{0}n^{1 }) – m(_{13}Al^{ 2}^{7})

=25.986895 +1.008665 -26.981541

=0.014019 u

=0.014019 × 931.5 MeV /c^{2}

Hence, the energy required for neutron removal is calculated as:

E =Δmc^{2}

=0.014019 ×931.5 = 13.059MeV

25. A source contains two phosphorous radio nuclides _{15}P^{32 }(T_{1/2} = 14.3d) and _{15}P^{33 }(T_{1/2} = 25.3d) Initially, 10% of the decays come from _{15}P^{33 .} How long one must wait until 90% do so?

Half life of _{15}P^{32} , T_{ 1/2} = 14.3 days

Half life of _{15}P^{33 }, p , T_{ 1/2 } = 25.3 days nucleus decay is10% of the total amount of decay

The source has initially 10% of _{15}P^{32 } nucleus and 90% of _{15}P^{33 }nucleus

Initially:

Number of _{15}P^{33 }nucleus = N

Number of _{15}P^{32 }nucleus = 9N

Finally:

Number of _{15}P^{33 }nucleus = 9N’

Number of _{15}P^{32 }nucleus = N’

For _{15}P^{32 } nucleus, we can write the number ratio as:

N’/9N =(1/2) ^{t/T1/2}

N’ =9N(2) ^{-t/14.3 ……(i)}

For _{15}P^{33} ,we can write the number ratio as:

9N’ /N = (1/2) ^{1/T ‘ 1/2 }

9N’ = N(2) ^{-t/25.3 ……..(ii)}

On dividing equation (1) by equation (2), we get:

1/9 = 9 ×2 ^{( t/25.3 – t/14.3)}

1/81 =2 ^{(11 t/25.3 × 14.3)}

log 1 – log 81 = (-11 t/ 25.3 × 14.3 ) log 1

-11 t/25.3 × 14.3 = o- 1.908/0.301

t = 25.3 × 14.3× 1.908 / 11 ×0.301

≈208.5 days

Hence, it will take about 208.5 days for 90% decay of _{15}P^{33}

26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

_{88}Ra^{ 223} → _{ 82}Pb ^{209} + _{ 6} C^{14}

_{88}Ra^{ 223} → _{ 86}Rn ^{219} + _{ 2}He^{4}

^{Calculate the Q-values for these decays and determine that both are energetically allowed}

Take a _{6} C^{14 } emission nuclear reaction:

_{88} Ra ^{223 } →_{ 82} Pb ^{209 }+ _{6} C^{14 }

We know that:

Mass of _{88} Ra ^{223 } ,m_{1} = 223.01850 u

Mass of _{82} Pb ^{209} , m_{2} = 208.98107 u

Mass of _{6} C^{14 }, m_{3} = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q= (m_{1} -m_{2} -m_{3})c^{2}

= (223.01850 – 208.98107 – 14.00324)c^{2}

=(0.03419 c^{2} ) u

but 1 u = 931.5 MeV /c^{2}

∴Q = 0.03419 931.5

= 31.848MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take a _{2}He^{4 } emission nuclear reaction:

_{88} Ra ^{223 } → _{86}Rn^{219} +_{2}He^{4}

We know that

Mass of _{88} Ra ^{223 } ,m_{1} = 223.01850

Mass of _{82}Rn^{219} ,m_{2} =219.00948

Mass of _{2}He^{4 } ,m_{3} = 4.00260

Q-value of this nuclear reaction is given as:

Q =(m_{1} -m_{2} -m_{3})c^{2}

=(223.01850 – 219.00948 – 4.00260 ) c^{2}

=(0.00642 c^{2}) u

=0.00642 × 931.5 = 5.98MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

27.Consider the fission of _{92} U ^{238} by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are _{58} Ce ^{140 } and _{ 44}Ru^{ 99} Calculate Q for this fission process. The relevant atomic and particle masses are

m(_{92} U ^{238} ) = 238.05079 u

m(_{58} Ce ^{140} ) = 139.90543 u

m(_{44}Ru^{ 99} )=98.90594 u

In the fission of _{92} U^{ 238} 10 β – particles decay from the parent nucleus. The nuclear reaction can be written as:

_{92} U^{ 238 }+ _{0}n^{1} → _{58}Ce ^{140 }+ _{44} Ru ^{99}+ 10 ^{0} _{-1} e

It is given that:

Mass of a nucleus m_{1} , _{92} U^{ 238 }= 238.05079 u

Mass of a nucleus m_{2} , _{58}Ce ^{140} = 139.90543 u

Mass of nucleus ,m_{3} _{44} Ru ^{99}= 98.90594 u

Mass of a neutron m_{4} _{0}n^{1} = 1.008665 u

Q-value of the above equation

Q= [ m’ ( _{92} U^{ 238}) + m (_{0}n^{1}) – m’ (_{58}Ce ^{140}) -m’ (_{44} Ru ^{99}) -10 m_{e}]c^{2}

Where,

m’ = Represents the corresponding atomic masses of the nuclei

m’ ( _{92} U^{ 238}) =m_{1} -92m_{e}

m’ ( _{58}Ce ^{140}) =m_{2} -58m_{e}

m’ ( _{44} Ru ^{99}) =m_{3} -44m_{e}

m(_{0}n^{1}) =m_{4}

Q= [ m_{1} -92m_{e }+ m_{4} -m_{2} +58m_{e} -m_{3} +44m_{e} -10m_{e}]c^{2}

=[ m_{1}+m_{4}– m_{2}-m_{3}]c^{2}

[238.0507 +1.008665 – 139.90543 – 98.90594]c^{2}

=[0.247995 c^{2} ]u

but 1 u = 931 MeV/C^{2}

∴Q=0.247995 ×931.5 =231.007MeV

Hence, the Q-value of the fission process is 231.007 MeV.

28.Consider the D−T reaction (deuterium−tritium fusion) _{1} H^{2} + _{1}H^{3} → _{2}He ^{4} +n

(a) Calculate the energy released in MeV in this reaction from the data

m( _{1} H^{2}) =2.014102 u

m(_{1}H^{3}) = 3.016049 u

b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kineticenergy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Solution 28:

(a) Take the D-T nuclear reaction: _{1}H^{2} + _{1}H^{3} →_{2}He^{4} +n

It is given that:

Mass of _{1}H^{2 }, m_{1} = 2.014102 u

Mass of _{1}H^{3}^{ }, m_{2} =3.016049 u

Mass of_{2}He^{4}^{ }, m_{3} = 4.002603 u

Mass of_{0}n^{1}^{ }, m_{3} = 1.008665 u

Q-value of the given D-T reaction is:

Q= [2.014102 + 3.016049 – 4.002603 – 1.008665] c^{2} = [0.018883 c^{2}]u

but 1 u = 931.5 MeV /c^{2}

∴Q= 0.018883 × 931.5 = 17.59 MeV

(b) Radius of deuterium and tritium, r ≈ 2.0 fm = 2 ×10 ^{-15} m

Distance between the two nuclei at the moment when they touch each other,

d= r+r = 4 × 10^{-15} m

Charge on the deuterium nucleus = e

Charge on the tritium nucleus = e

Hence, the repulsive potential energy between the two nuclei is given as:

V = e^{2} / 4πε_{0} (d)

Where,

ε_{0 }= Permittivity of free space

1/4πε_{0 } = 9 ×10 ^{9 }Nm^{2} c^{-2} lim _{x→∞}

∴ V =9 × 10^{9} × (1.6 × 10 ^{-19} ) ^{2 } / 4 ×10 ^{-15 } = 5.76 × 10 ^{-14} J

= 5.76 × 10 ^{-14 }/ 1.6 × 10 ^{-19 } = 3.6 × 10 ^{5 }eV =360keV

Hence, 5.76 × 10 ^{-14 }J or r keV of kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.

However, it is given that

KE= 2 ×(3K T/2)

Where,

k =Boltzmann constant = 1.38 × 10 ^{-23} m^{2} kg s^{-2} k^{-1}

T= Temperature required for triggering the reaction

∴ T =KE/3K

= 5.76 ×10 ^{-14 }/ 3 × 1.38 × 10 ^{-23 }=1.39 ×10 ^{9 }K

Hence, the gas must be heated to a temperature of 1.39 ×10 ^{9 }Kto initiate the reaction.

29.Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ – decays in the decay scheme shown in figure. You are given that

m(^{198} Au) =197.968233u

m(^{198} Hg) =197.966760 u

It can be observed from the given γ -decay diagram that γ _{1 }decays from the 1.088 MeVenergy level to the 0 MeV energy level

Hence, the energy corresponding to γ_{ 1} -decay is given as:

E_{1} = 1.088 – 8 =1.088 MeV hv _{1}= 1.088 ×1.6 × 10^{-19 } × 10 ^{6}J

Where,

h =Planck’sconstant = 6.6 × 10 ^{-34 }JSV_{1 }= Frequency of radiation radiated by γ_{ 1} decay

∴ v_{1 }= E_{1}/h

= 1.088 × 1.6 ×10 ^{-19 } × 10 ^{6 }/ 6.6 × 10 ^{-34 }= 2.637 × 10 ^{20 }Hz

It can be observed from the given γ-decay diagram that γ_{2 }decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ_{2 } -decay is given as:

E_{2 }= 0.412 – 0 = 0.412MeV hv^{2}=0.412 ×1.6 ×10 ^{-19 } ×10^{6 J}

Where,

v_{2 }= Frequency of radiation radiated by γ_{2 – decay}

∴ v_{2} = E_{2}/h

=0.412 × 1.6 × 10 ^{-19 } × 10^{6 }/ 6.6 × 10 ^{-34 } = 9.988 ×10 ^{19 }Hz

It can be observed from the given γ -decay diagram that γ_{3} decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to γ_{3} -decay is given as:

E_{3} = 1.088 -0.412 =0.676 MeV hv^{3 } = 0.676 × 10 ^{-19 } ×10 ^{6}

where,

v_{3 }= Frequency of radiation radiated by γ_{3 decay}

∴ v_{3 }= E_{3}/h

= 0.676×1.6 ×10 ^{-19 } × 10 ^{6 }/6.6 × 10 ^{-34 } = 1.639 × 10 ^{20 }Hz

Mass of m (_{78} Au ^{198}) = 197 .968233 u

Mass of m( _{80} Hg ^{198}) = 197.966760 u

1 u = 931 .5 MeV /c^{2}

Energy of the highest level is given as:

E= [m((_{78} Au ^{198}) – m ( _{80} Hg ^{198}) ]

= 197.968233 – 197.966760 = 0.001473 u

= 0.001473 × 931.5 = 1.3720995MeV

β_{1} decays from the 1.3720995 MeV level to the 1.088 MeV level

Maximum kinetic energy of the β_{1 } particle = 1.3720995 − 1.088

= 0.2840995 MeV

β _{2 }decays from the 1.3720995 MeV level to the 0.412 MeV level

Maximum kinetic energy of the β _{2 }particle = 1.3720995 – 0.412

= 0.9600995 MeV

30.Calculate and compare the energy released by

(a) fusion of 1.0 kg of hydrogen deep within Sun and

(b) the fission of 1.0 -kg ^{235} U in a fission reactor.

(a) Amount of hydrogen, m = 1 kg = 1000 g

1 mole, i.e., 1 g of hydrogen (_{1}H^{1}) contains 6.023 × 10^{23} Atoms

∴ 1000 g of _{1}H^{1} contains 6.023 × 10 ^{23 }× 1000 atoms

Within the sun, four _{1}H^{1 } nuclei combine and form one _{2}He^{4 }nucleus. In this process 26MeV of energy is released.

Hence, the energy released from the fusion of 1 kg _{1}H^{1 is }

E_{1 }=6.023 ×10^{23} ×26 × 10 ^{3 }/4

=39.1495 × 10 ^{26 }MeV

b)Amount of _{92} U ^{235 } =1000gm

1 mole, i. e., ,235 g of _{92} U ^{235 }contains 6.023 × 10 ^{23 }atoms

∴ 1000g of _{92} U ^{235 } contains 6.023 ×10^{23} × 1000 /235 atoms

It is known that the amount of energy released in the fission of one atom of _{92} U ^{235 } 200 MeV

Hence, energy released from the fission of 1kg of _{92} U ^{235 }is

E_{2 } = 6 × 10 ^{23} × 1000 ×200 /235

= 5.106 ×10^{26 } MeV

E^{1}/E_{2} = 39.1495 × 10 ^{26}/ 5.106 × 10 ^{26 }=7.67 ≈ 8

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium

31. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of^{ 235}U to be about 200MeV.

Amount of electric power to be generated, p =2 × 10 ^{5 } MW 10% of this amount has to be obtained from nuclear power plants.

Amount of nuclear power, p_{1 }=( 10 /100 ) × 2 × 10 ^{5}

= 2× 10 ^{4 }MW

=2 × 10^{ 4 } ×10 ^{6 } J/s

= 2 ×10 ^{10 }× 60 ×60 ×24 ×365 J/y

Heat energy released per fission of a _{235} U nucleus, E =200 MeV

Efficiency of a reactor =25%

Hence, the amount of energy converted into the electrical energy per fission is calculated as:

(25/100 ) 200 = 50 MeV

=50 × 1.6 ×10 ^{-19 } × 10 ^{6 }=8 ×10 ^{-12 }J

Number of atoms required for fission per year:

2 × 10^{10} ×60 ×60×24 ×365 / 8 ×10 ^{-12 }= 78840 × 10 ^{24 } atoms

1 mole, i.e., 235g of U ^{235} contains 6.023 × 10 ^{23} Atoms

∴Mass of 6.023 × 10 ^{23 }atoms of U ^{235} =235g = 235 × 10 ^{-3} kg

∴ Mass of 78840 × 10 ^{24} atoms of U ^{235}

= (235 × 10 ^{-3 }/ 6.023 ×10 ^{23} ) 78840 × 10 ^{24}

= 3.076 × 10 ^{4 }kg

Hence, the mass of uranium needed per year is 3.076 × 10 ^{4 }kg